Rectangle $ABCD$ is inscribed in triangle $EFG$ such that side $AD$ of the rectangle is on side $EG$ of the triangle, as shown. The triangle's altitude from $F$ to side $EG$ is 7 inches, and $EG = 10 \text{ inches}$. The length of segment $AB$ is equal to half the length of segment $AD$. What is the area of rectangle $ABCD$? Express your answer as a common fraction.

[asy]
import math;
size(101);
real x = 35/12;
currentpen = linewidth(1)+fontsize(10pt);
pair E1 = (0,0), G = (10,0), F = (3,7), A = (3*x/7,0), D = G - (7*x/7,0), B = extension(E1,F,A,A+(0,1)), C = extension(G,F,D,D+(0,1));
draw(E1--F--G--cycle); draw(A--B--C--D); label("$A$",A,S); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,S); label("$E$",E1,W); label("$F$",F,NNW); label("$G$",G,ESE);
[/asy]
Suppose that the altitude from $F$ to $EG$ intersects $EG$ at point $H$.  Then $\triangle EAB \sim \triangle EHF$, and we have that $\frac{HE}{HF} = \frac{AE}{AB}$.  Also, $\triangle GDC \sim GHF$, and $\frac{HG}{HF} = \frac{DG}{DC}$.  Adding these equalities, we find that $\frac{HE + HG}{HF} = \frac{AE + DG}{AB}$, since $AB = DC$.  But $HE + HG = EG = 10$, $HF = 7$, and finally $AE + DG = EG - AD = 10 - 2AB$.  Plugging in, we find that $\frac{10}{7} = \frac{10-2AB}{AB}$, or $AB = \frac{35}{12}$.  Thus the area of $ABCD$ is $\frac{35}{12}\cdot\frac{35}{6} =\boxed{ \frac{1225}{72}}$.